MCQ Questions for Class 7 Maths Perimeter And Area Quiz 9

Take a circle with radius

$5$ m, then determine the area of given circle with suitable units.

0%
$78.53\ m$
0%
$78.53\ m^{2}$
0%
$78.53\ cm$
0%
$78.53\ cm^{2}$

Explanation

Area of circle

$=\pi { r }^{ 2 }=\cfrac { 22 }{ 7 } \times 5\times 5=78.57m^2$

Find the circumference of a circle whose radius is-

$35\ cm$

0%
$220\ cm$
0%
$20\ cm$
0%
$22\ cm$
0%
None of these

Explanation

Given: Radius of circle

$(r)=35\space\mathrm{cm}$

$\therefore$ Circumference of circle

$=2\pi r=2\times\dfrac{22}{7}\times35=220\space\mathrm{cm}$

So,

$\text{A}$ is the correct option.

If the area of a semicircle is 84 cm

$^2$ , then the area of the circle is .......................

0%
144 cm $^2$
0%
42 cm $^2$
0%
168 cm $^2$
0%
288 cm $^2$

Explanation

We know that the area of semicircle is one half of a circle .

Given area of semicircle is

$84 cm^{2}$

Then area of circle

$=2\times 84$

$=168 cm^{2}$

Calculate perimeter of a right angled triangle whose non-hypotenuse sides are

$6$ cm and

$8$ cm.

0%
$18\ cm$
0%
$14\ cm$
0%
$24\ cm$
0%
$11\ cm$

Explanation

In right

$\triangle PQR$ , using Pythagoras theorem,

$P{ R }^{ 2 }=P{ Q }^{ 2 }+Q{ R }^{ 2 }={ 6 }^{ 2 }+{ 8 }^{ 2 }=36+64=100$

$PR=\sqrt { 100 } =10㎝$

$\therefore$ Perimeter of

$PQR=PQ+QR+RP=6+8+10=24 cm$

In a parallelogram the base and height are is in the ratio of

$5 : 2$ . If the area of the parallelogram is

$360\ m^{2}$ , find its base and height.

0%
$30\ m, 12\ m$ .
0%
$20\ m, 12\ m$ .
0%
$30\ m, 10\ m$ .
0%
None of these

Explanation

The ration of the base and height of the parallelogram is 5 : 2

$\therefore$ let, the base and height be 5x & 2x respectively.

We know, area of a parallelogram is given by,

$\dfrac{1}{2} \textrm{base} \times \textrm{height}$ .

$\therefore$ according to the question,

$\dfrac{1}{2} \times {5x} \times {2} = 360$ [ The area is given by 360

$m^2$ ]

$\Rightarrow x^2 = 36$

$\Rightarrow x = 6 [ \therefore$ base & height cannot be negative ]

$\therefore$ The base is ( 5 X 6 ) = 30 m . and the height is ( 2 X 6 ) = 12 m .

$\therefore$ option Ais correct.

In the given figure, each square contains four square grid of equal dimensions. Then calculate the area of the given figure.

0%
$16\ cm^{2}$
0%
$20\ cm^{2}$
0%
$24\ cm^{2}$
0%
$28\ cm^{2}$

Explanation

Area of figure

$=4\times$ area of

$1$ square

$=4{ a }^{ 2 }$

Put

$a=2$ , then area

$=4\times { 2 }^{ 2 }=16 cm^2$

In Figure

$ABCD$ is a square then find the area of the shaded region.

0%
$800\ cm^{2}$ .
0%
$600\ cm^{2}$ .
0%
$200\ cm^{2}$ .
0%
None of these

Explanation

to find out the area of shaded region let us follow the method;

Area of shaded region

$=$ Area of asquare

$ABCD-$ Area of triangle

$ABE$

Area of square of side

$a$ is equal to

$a^2$

for

$a=40cm$

$\therefore$ Area of square

$ABCD=40 \times 40=1600cm^2$

Area of triangle

$ABE=\dfrac{1}{2}\times$ base

$\times$ height

For base

$=40cm;height=40cm$

Area of triangle

$ABE=\dfrac{1}{2}\times 20\times 40\Rightarrow 20\times 40 sqcm$

$\therefore$ Area of triangle

$\Rightarrow 800cn^2$

Now, Area of shaded region

$=$ area of square

$ABCD -$ area of triangle

$ABE$

$\Rightarrow 1600cm^2-800cm^2=800cm^2$

$\therefore$ Area of shaded region

$=800cm^2$

1148098_703311_ans_e8d6a2f8da564fa89b2220a0d64937a7.png

If the circumference of a circle is

$264\ cm$ , find its radius. Take

$\pi = \dfrac {22}{7}$ .

0%
$42\ cm$ .
0%
$22\ cm$ .
0%
$32\ cm$ .
0%
None of these

Explanation

Let the radius be

$r\ cm$

Given Circumference of a circle is

$264\ cm$

$\implies 2\pi r=264$

$\implies 2\times \dfrac{22}{7}\times r=264$

$\implies r=\dfrac{264\times 7}{44}=42$

So the radius is

$42\ cm$

Rohit's family is planning to build a concrete basketball court in their backyard. How many square metres will be left in the backyard for grass?

0%
$60$ sq. cm
0%
$35$ sq. cm
0%
$31$ sq. cm
0%
$24$ sq. cm

Explanation

$1$ sq. m

$=1\square$
Area of the concrete basketball court

$=7\times 5=35$ sq. m
Area of the backyard

$=11\times 6=66$ sq. m
Area of the backyard left for grass

$=(66-35)$ sq. m

$=31$ sq. m.

Perimeter of circle is known as

0%
Area
0%
Volume
0%
Diameter
0%
Circumference

If the circumference of a circle is

$60$ cm more than its diameter, then the circumference is?

0%
$14\pi$ cm
0%
$28\pi$ cm
0%
$35\pi$ cm
0%
$42\pi$ cm

Explanation

Given: Circumference of circle

$=60+$ Diameter

$\Rightarrow 2\pi r=60+2r$

$\Rightarrow$

$2\times \displaystyle\frac{22}{7}\times r=60+2r$

$\Rightarrow$

$\displaystyle\frac{22}{7}\times r=30+r$

$\Rightarrow$

$30=\displaystyle\frac{22}{7}r-r$

$\Rightarrow$

$30=\displaystyle\frac{15r}{7}$

$\Rightarrow r=14$

Hence circumference

$=2\pi r$

$=2\times \dfrac{22}{7} \times 14$

$=88$

The shaded part in the given figure is covered with cement. If its costs Rs.

$84$ to cement an area of

$3$

$cm^2$ , find the total cost of cementing.

0%
Rs. $8820$
0%
Rs. $8125$
0%
Rs. $8610$
0%
Rs. $8804$

Explanation

Area of larger rectangle

$=(32\times 25)cm^2=800cm^2$
Also, area of smaller rectangle

$=485cm^2$

$\therefore$ Area of shaded part

$=(800-485)cm^2=315cm^2$
Now,
Cost of cementing of

$3cm^2=$ Rs.

$84$

$\therefore$ Cost of cementing of

$1cm^2=(84\div 3)=$ Rs.

$28$

$\therefore$ Cost of cementing of

$315cm^2$

$=$ Rs.

$(28\times 315)=$ Rs.

$8820$ .

If the area of a rectangle is equal to the area of a square and if one side (

$l$ ) of the rectangle is equal to the perimeter of the square, then the other side (

$b$ ) of rectangle is _______ .

0%
$side\div 2$
0%
${side}^{2}\div 2$
0%
$side \div 4$
0%
$side\div 3$

Explanation

Perimeter of square

$=4\times$ side

Then,

$l=4\times$ side

Area of rectangle=Area of square

$\Rightarrow l\times b=$ side

$\times$ side

$\Rightarrow \left( 4\times side \right) \times b=$ side

$\times$ side

$\Rightarrow b=\cfrac { side\times side }{ 4\times side } =\cfrac { side }{ 4 }$

If the circumference of a circle is

$33\ cm$ , find its diameter.

0%
$10.5\ cm$ .
0%
$12.5\ cm$ .
0%
$16.5\ cm$ .
0%
None of these

Explanation

Given: Circumference of circle

$=33\space\mathrm{cm}$

$\Rightarrow 2\pi r=33$

$\Rightarrow 2\times\dfrac{22}{7}\times r=33$

$\Rightarrow r=33\times\dfrac{1}{2}\times\dfrac{7}{22}$

$\Rightarrow r=5.25\space\mathrm{cm}$

$\Rightarrow d=2r=2\times5.25=10.5\space\mathrm{cm}$

$\therefore$ Diameter of circle

$d=10.5\space\mathrm{cm}$

Hence.

$\text{A}$ is the correct option.

Find the circumference of circle whose diameter is-

$17.5\ cm$
Take

$\pi = \dfrac {22}{7}$ in the above two questions.

0%
$55\ m$
0%
$25\ m$
0%
$59\ m$
0%
None of these

Explanation

Given: Diameter of circle

$(d)=17.5\space\mathrm{m}$

$\Rightarrow$ Radius of circle

$(r)=\dfrac{17.5}{2}=8.75\space\mathrm{m}$

Let

$C$ be the circumference of circle,

$C=2\pi r\\=2\times\dfrac{22}{7}\times8.75\\=55\space\mathrm{m}$

So,

$\text{A}$ is the correct option.

The length of a rectangle is

$\left( \cfrac { 6 }{ 5 } \right)$ th of its breadth. It its perimeter is

$132m$ , its area will be ______ .

0%
$1080{m}^{2}$
0%
$640{m}^{2}$
0%
$1620{m}^{2}$
0%
$2160{m}^{2}$

Explanation

$\Rightarrow$ Let breadth of a rectangle be

$x$ .

$\Rightarrow$ Then length of a rectangle be

$\dfrac{6}{5}x$ .

$\Rightarrow$ Perimeter of a rectangle =

$2(length+breadth)$

$\therefore$

$132=2(\dfrac{6}{5}x+x)$

$\therefore$

$66=\dfrac{6+5}{5}x$

$\therefore$

$x=66\times \dfrac{5}{11}$

$\therefore$

$x=30\,m$

$\Rightarrow$ So, breadth =

$30\,m$ and length =

$\dfrac{6}{5}\times 30=36\,m$

$\Rightarrow$ Area of rectangle =

$length\times breadth$

$\therefore$ Area of rectangle =

$36\times 30=1080\,m^2$

The area of the shaded region in the figure is ______ .

0%
$207\ \text{m}^{2}$
0%
$158\ \text{m}^{2}$
0%
$256\ \text{m}^{2}$
0%
$60\ \text{m}^{2}$

Explanation

$\Rightarrow$ We can see, given figure is a square. There are

$4$ equal small squares inside of large square.

$\Rightarrow$ Side of large square is

$16\,m$

$\therefore$ Area of large square =

$16^2=256\,m^2$

$\Rightarrow$ Side of small square is

$7\,m$ , there are

$4$ small equal squares.

$\therefore$ Area of small

$4$ squares =

$4\times 7^2=196\,m^2$

$\therefore$ The area of the shaded region = Area of large square - Area of

$4$ small squares.

$\therefore$ The area of the shaded region =

$256\,m^2-196m^2=60\,m^2$

Ishika has designed a small oval race track for her remote control car. Her design is shown in the figure. What is the total distance around the track? Round your answer to the nearest whole cm.

0%
$205\ cm$
0%
$305\ cm$
0%
$100\ cm$
0%
$245\ cm$

Explanation

Total distance around the track

$=$ Perimeter of the given track

The track is a combination of a rectangular path and two semi-circular paths.

Hence perimeter of the given track

$= 52 + 52 + 2\pi r$

$= 52 + 52 + 2\times \dfrac {22}{7}\times 16$

$= 104 + 100.57$

$= 204.57 \approx 205\ cm$ .

The area of a square is numerically equal to the perimeter of the square, then the side of square is ______ .

0%
$2$ units
0%
$3$ units
0%
$4$ units
0%
$5$ units

Explanation

$\Rightarrow$ Let

$'s'$ be the side of the square.

$\rightarrow$ According to the given condition,

$\Rightarrow$

$Area\,of\,square=Perimeter\,of\,square$

$\Rightarrow$

$s^2=4s$

$\therefore$

$s=4.$

$\therefore$ Side of square is

$4\,units.$

The number of envelopes that can be made out of sheet of paper

$384cm$ by

$172cm$ if each envelope requires a piece of paper of size

$16cm$ by

$12cm$ , is _________ .

0%
$340$
0%
$344$
0%
$338$
0%
$342$

Explanation

$\Rightarrow$ Length and breadth of a sheet is

$384\,cm$ and

$172\,cm$ .

$\Rightarrow$ Area of a sheet =

$length\times breadth$

$\therefore$ Area of a sheet =

$384\,cm\times 172\,cm=66048\,cm^2$

$\Rightarrow$ Length of a envelope is

$16\,cm$ and breadth is

$12\,cm$

$\Rightarrow$ Area of envelope =

$length\times breadth$

$\therefore$ Area of a envelope =

$16\,cm\times 12\,cm=192\,cm^2$

$\Rightarrow$

$Number\, of\, envelopes = \dfrac{Area\,of\,sheet}{Area\,of\,envelope}$

$\therefore$

$Number\, of\, envelopes =\dfrac{66048}{192}=344$

A square and a rectangular plot of land have the same perimeter. If the square is of side

$40m$ and the rectangle is of length

$5$ decameter. Then area of rectangle is _______ .

0%
$1500{m}^{2}$
0%
$1600{m}^{2}$
0%
$200{m}^{2}$
0%
$150{m}^{2}$

Explanation

$\Rightarrow$ Length of a rectangle

$=5$ decameter

$=50\,m$

$\Rightarrow$ Side of square =

$40\,m$

$\therefore$ Perimeter of squares =

$4\times 40=160\,m$

$\Rightarrow$ It is given that, perimeter of square = perimeter of a rectangle.

$\Rightarrow$ Perimeter of rectangle =

$2(l+b)$

$\therefore$

$160=2(50+b)$

$\therefore$

$80=50+b$

$\therefore$

$b=30\,m$

$\Rightarrow$ Area of a rectangle

$=l\times b=50\times 30=1500\,m^2$

All the figures given below consist of the four squares of equal size. Which figure has the smallest perimeter?

Explanation

$\Rightarrow$ We have given that Four squares in diagram has equal size.

$\Rightarrow$ Let the each square be

$1\times 1$ unit square.

$\Rightarrow$ Perimeter of Option A diagram =

$1+1+1+1+1+1+1+1=8$

$\Rightarrow$ Perimeter of Option B diagram =

$1+1+1+1+1+1+1+1+1+1=10$

$\Rightarrow$ Perimeter of Option C diagram =

$1+1+1+1+1+1+1+1+1+1=10$

$\Rightarrow$ Perimeter of Option D diagram =

$1+1+1+1+1+1+1+1+1+1=10$

$\therefore$ Option

$A$ diagram has the smallest perimeter.

What is the perimeter of the shaded area in the given figure?

0%
$18$ units
0%
$19$ units
0%
$16$ units
0%
$9$ units

Of the following statements, the one that is incorrect is:

0%
Doubling the length of a given rectangle doubles the area.
0%
Doubling the altitude of a triangle doubles the area.
0%
Doubling the radius of a given circle doubles the area .
0%
Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.

Explanation

We know, Area of circle

$A=\pi r^2$

When the radius is doubled,

$R=2r$

Then, New area

$=\pi R^2$

$=\pi (2r)^2$

$=4A$

Area of a circle is directly proportional to square of its radius, so doubling radius will quadruple its area.

If a bicycle wheel makes

$5000$ revolution in moving

$11$ km, then diameter of wheel is

0%
$110$ cm
0%
$120$ cm
0%
$70$ cm
0%
$100$ cm

Explanation

Let radius of wheel be

$r㎝$ . Then, distance travel in

$1$ revolution

$=2\pi r$

Total distance

$=11㎞=1100m=1100000㎝$

Total revolutions

$=5000$

$\Rightarrow 5000\times 2\pi r=1100000\Rightarrow r=\cfrac { 1100000\times 7 }{ 2\times 22\times 5000 } =35㎝$

$\therefore$ Diameter of wheel

$=2\times 35=70㎝$

Two concentric circles form a ring. The inner and outer circumferences of the ring are

$50\dfrac{2}{7}\ m$ and

$75\dfrac{3}{7}\ m$ respectively. Find the width of the ring.

0%
$3\ m$
0%
$4\ m$
0%
$5\ m$
0%
$6\ m$

If the base and altitude of a parallelogram are doubled, what happens to the area compared to the original one?

0%
$8$ times original
0%
$2$ times original
0%
$4$ times original
0%
Remains same

Explanation

Area of parallelogram

$S_1=bh$ , where

$b$ is the base and

$h$ is the height.

New Area,

$S_2=2b\times 2h$

$=4bh$

$=4S_1$

Hence, the area becomes

$4$ times the original area.

Find the area of a parallelogram having base

$4$ cm and height

$2$ cm.

0%
$6$ sq. cm
0%
$8$ sq. cm
0%
$10$ sq. cm
0%
$12$ sq. cm

Explanation

Base of the parallelogram,

$b=4$ cm

Height,

$h=2$ cm

Area of the parallelogram

$=b\times h$

$=4\times 2$

$=8$

$cm^{2}$

1033406_1069485_ans_92021c967f44400fae369780653949e3.jpg

Find the area of the circular park of Janakpuri whose circumference is

77m77m

0%
$671.87m^2$
0%
$331.62m^2$
0%
$471.62m^2$
0%
$401.62m^2$

Explanation

Given that,

Circumference of a circle $=77\,m$

We know that,

Circumference of a circle $=2\pi r=77$

$\Rightarrow 2\times \dfrac{22}{7}\times r=77$

$\Rightarrow r=\dfrac{77\times 7}{2\times 22}$

$\Rightarrow r=\dfrac{49}{4}$

Then, Area of circular park

$=\pi {{r}^{2}}=\dfrac{22}{7}\times \dfrac{49}{4}\times \dfrac{49}{4}$

$\Rightarrow \dfrac{11\times 7\times 49}{2\times 4}$

$\Rightarrow \dfrac{77\times 49}{8}$

$\Rightarrow \dfrac{3773}{8}\,\,{{m}^{2}}$

$or$

$\Rightarrow 471.62\,{{m}^{2}}$

The circumference of a circle is

$31.4\ cm$ . Find the area of the circle? (Take

$\pi=3.14$ )

0%
$58.5cm^2$
0%
$48.5cm^2$
0%
$78.5cm^2$
0%
$88.5cm^2$

Explanation

Given,

circumference of the circle

$= 31.4cm$

Let, the radius be

$r$

$\Rightarrow 2\pi r = 31.4$

$\Rightarrow r = \dfrac{31.4}{2\times \pi}$

$\Rightarrow r = \dfrac{31.4}{2\times 3.14}$

$\Rightarrow r = 5$

So, the radius

$r=5\ cm$

Now, Area of circle

$= \pi r^2$

$= 3.14\times (5)^2\ cm^2$

$= 3.14\times 25\ cm^2$

$=78.5\ cm^2$

Hence, the area of the circle is

$78.5\ cm^2$ .

CBSE Questions for Class 7 Maths Perimeter And Area Quiz 9 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Perimeter And Area Quiz 9 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

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